Use Newton's method to approximate the positive value of x which satisfies x=2.3cosx

Let x0=1 be the initial approximation. Find the next two approximations, x_1 and x2, to four decimal places each.

A fine online calculator for Newton's Method is at

http://www.math.sc.edu/cgi-bin/sumcgi/Newton.pl

To use Newton's method to approximate the positive value of x that satisfies the equation x = 2.3cos(x), we need to repeatedly apply the following formula:

x_n+1 = x_n - f(x_n)/f'(x_n)

where x_n is the nth approximation, f(x_n) is the value of the function at x_n, and f'(x_n) is the derivative of the function at x_n.

Given the initial approximation x_0 = 1, we can calculate the next approximations x_1 and x_2 as follows:

1. Calculate f(x_0):
f(x_0) = 2.3*cos(x_0)

2. Calculate f'(x_0):
f'(x_0) = -2.3*sin(x_0)

3. Calculate x_1 using the formula:
x_1 = x_0 - f(x_0)/f'(x_0)

4. Calculate f(x_1):
f(x_1) = 2.3*cos(x_1)

5. Calculate f'(x_1):
f'(x_1) = -2.3*sin(x_1)

6. Calculate x_2 using the formula:
x_2 = x_1 - f(x_1)/f'(x_1)

Let's calculate x_1 and x_2 to four decimal places using Newton's method:

Step 1:
f(x_0) = 2.3*cos(1) = 1.38177329

Step 2:
f'(x_0) = -2.3*sin(1) = -1.66996833

Step 3:
x_1 = 1 - 1.38177329 / -1.66996833 = 1.82802

Step 4:
f(x_1) = 2.3*cos(1.82802) = 0.81895625

Step 5:
f'(x_1) = -2.3*sin(1.82802) = -1.76860039

Step 6:
x_2 = 1.82802 - 0.81895625 / -1.76860039 = 1.3077

Therefore, the next two approximations to four decimal places are:
x_1 ≈ 1.8280
x_2 ≈ 1.3077